If \( x=3\sin\theta-\sin3\theta,\ y=3\cos\theta-\cos3\theta \), then \( \displaystyle\frac{d^{2}y}{dx^{2}} \) at \( \theta=\displaystyle\frac{\pi}{3} \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( -\displaystyle\frac{16}{27} \)
Since x and y are given in terms of \( \theta \), differentiate each with respect to \( \theta \):
\( \displaystyle \frac{dx}{d\theta}=3(\cos\theta-\cos3\theta) \)
\( =6\sin2\theta\sin\theta \)
\( \displaystyle \frac{dy}{d\theta}=3(\sin3\theta-\sin\theta) \)
\( =6\cos2\theta\sin\theta \)
So
\( \displaystyle \frac{dy}{dx}=\cot2\theta \)
Differentiating again with respect to x,
\( \displaystyle \frac{d^{2}y}{dx^{2}}=\frac{-2\,\mathrm{cosec}^{2}2\theta}{dx/d\theta} \)
\( \displaystyle =\frac{-2\,\mathrm{cosec}^{2}2\theta}{6\sin2\theta\sin\theta} \)
At \( \theta=\frac{\pi}{3} \): \( \sin2\theta=\sin\theta=\frac{\sqrt3}{2} \) and \( \mathrm{cosec}^{2}2\theta=\frac{4}{3} \).
\( \displaystyle \frac{d^{2}y}{dx^{2}}=\frac{-2\cdot\frac{4}{3}}{6\cdot\frac{3}{4}} \)
\( \displaystyle =\frac{-\frac{8}{3}}{\frac{9}{2}} \)
\( \displaystyle =-\frac{16}{27} \)